I have a constrained maximisation problem (problem 1 below) and I would like your help to understand whether it is equivalent to problem 2. My intuition is that they are equivalent but I don't know how to prove it and hence I'm not sure.
Notation:
$\mathcal{I}\equiv \{1,2,...,I\}$, $\mathcal{J}\equiv \{1,2,...,J\}$
Unknowns: $x_{ij}$ $\forall (i,j)\in \mathcal{I}\times \mathcal{J}$, $x_{i0}$ $\forall i\in \mathcal{I}$, and $x_{0j}$ $\forall j \in \mathcal{J}$
$x$ denotes the vector of all unknowns of dimension $IJ+I+J$
$\tilde{x}\equiv (x_{ij}$ $\forall (i,j)\in \mathcal{I}\times \mathcal{J})$
Known parameters: $A_{ij}$ $\forall (i,j)\in \mathcal{I}\times \mathcal{J}$, $A_{i0}$ $\forall i\in \mathcal{I}$, and $A_{0j}$ $\forall j \in \mathcal{J}$
Problem 1:
$$ \max_x\Big[ \sum_{i=1}^I x_{i0}A_{i0}+\sum_{j=1}^J x_{0j}A_{0j}+\sum_{i=1}^I \sum_{j=1}^J x_{ij} A_{ij}\Big] $$ $$ \text{ s.t. } \sum_{j=0}^J x_{ij}=1 \text{ }\forall i \in \mathcal{I}\text{, }\sum_{i=0}^I x_{ij}=1 \text{ }\forall j \in \mathcal{J}\text{, } x\in \{0,1\}^{IJ+I+J} $$
Problem 2:
$$ \max_\tilde{x}\Big[ \sum_{i=1}^I \sum_{j=1}^J x_{ij} (A_{ij}-A_{i0}-A_{0j})\Big] $$ $$ \text{ s.t. } \sum_{j=1}^J x_{ij}\leq 1 \text{ }\forall i \in \mathcal{I}\text{, }\sum_{i=1}^I x_{ij}\leq 1 \text{ }\forall j \in \mathcal{J}\text{, } \tilde{x}\in \{0,1\}^{IJ} $$
Any hint?